Could any one please help me with interpretation of the following Zn-Imidazole titiration plot?
Two solutions: pure imidazole and imidazole with Zn(NO3)2 (mole ratio 10:1) were titrated with NaOH base. Both solutions contained excess HNO3 at the beginning.
Both plots were offset to help interpretation and 0 on x-axis means no free HNO3 exists in the solution.
The interpretation of pure imidazole plot is straightforward - first equivalence point at NaOH/Imidazole = 0 is equivalence point of free HNO3 and at NaOH/Im = 1 is equivalence of the protonated imidazole.
I'm not sure if I interpret the second (Im-Zn 10:1) plot correctly. I assume that red line lying (Zn-Im) lying below pure Im line means that complexation occurs. But why second equivalence point moved right? It looks like protonated imidazole and Zn are titrated separately, because I needed 1.2 moles of base. I assume that following amounts of base are used: 1 mole per 1 mole of imidazole and 0.2 moles for Zn, to form Zn(OH)2, per 0.1 moles of Zn - at the beginning I had ten times less Zn ions than imidazole.
Additionally at pH = 7.3 precipitation occurred. Are there any clues if the precipitate contains Zn(OH)2 or maybe insoluble Zn-Im complex? And finally are there any direct proofs on the diagram that complexation occurs?
Thank you Michael. I will definitely try additional analysis of the precipitate, as you suggested.
Interpreting the plot is relatively simple, and the original post has most of the answers already. The red curve is below that of pure Im because there is complexation of Zn although the complex formed seems rather weak. The equivalence point moved right because there were additional deprotonation equilibria taking place compared to that of pure Im. Whether Im and Zn were titrated separately or as a complex is not possible to know without other analysis. Either the complex is so weak that around neutral pH it dissociates to form Zn(OH)2 which precipitates, or at that point 2 water molecules coordinated to Zn deprotonate and a hydroxocomplex species of the form Zn(Im)x(OH)2 is obtained as an insoluble compound. In any case, there is definitely proof of complex formation above pH=5. And the precipitate must contain Zn, alone or in an Im complex species. Precipitate analysis can tell which one is correct.
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